Find the possible values of 'a' such that the inequality `3-x^(2)gt |x-a|` has atleast one negative solution

Answer

To solve the inequality \( 3 - x^2 > |x - a| \) and find the possible values of \( a \) such that there is at least one negative solution, we can follow these steps:

### Step 1: Analyze the inequality
We start by rewriting the inequality:
\[ 3 - x^2 > |x - a| \]
This can be split into two cases based on the definition of absolute value.

### Step 2: Case 1: \( x - a \geq 0 \) (i.e., \( x \geq a \))
In this case, the inequality becomes:
\[ 3 - x^2 > x - a \]
Rearranging gives:
\[ 3 + a > x + x^2 \]
This can be rewritten as:
\[ x^2 + x - (3 + a) < 0 \]

### Step 3: Find the roots of the quadratic
To find the roots of the quadratic equation \( x^2 + x - (3 + a) = 0 \), we use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1, b = 1, c = -(3 + a) \):
\[ x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot -(3 + a)}}{2 \cdot 1} \]
\[ x = \frac{-1 \pm \sqrt{1 + 12 + 4a}}{2} \]
\[ x = \frac{-1 \pm \sqrt{13 + 4a}}{2} \]

### Step 4: Condition for the quadratic to be negative
For the quadratic \( x^2 + x - (3 + a) < 0 \) to have real roots, the discriminant must be non-negative:
\[ 13 + 4a \geq 0 \]
This gives:
\[ a \geq -\frac{13}{4} \]

### Step 5: Case 2: \( x - a < 0 \) (i.e., \( x < a \))
In this case, the inequality becomes:
\[ 3 - x^2 > -(x - a) \]
Rearranging gives:
\[ 3 - x^2 + x - a > 0 \]
This can be rewritten as:
\[ -x^2 + x + (3 - a) > 0 \]
or
\[ x^2 - x - (3 - a) < 0 \]

### Step 6: Find the roots of the second quadratic
Using the quadratic formula again:
\[ x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 1 \cdot -(3 - a)}}{2 \cdot 1} \]
\[ x = \frac{1 \pm \sqrt{1 + 12 - 4a}}{2} \]
\[ x = \frac{1 \pm \sqrt{13 - 4a}}{2} \]

### Step 7: Condition for the second quadratic to be negative
For the quadratic \( x^2 - x - (3 - a) < 0 \) to have real roots, the discriminant must also be non-negative:
\[ 13 - 4a \geq 0 \]
This gives:
\[ a \leq \frac{13}{4} \]

### Step 8: Combine the conditions
From the two cases, we have:
1. \( a \geq -\frac{13}{4} \)
2. \( a \leq \frac{13}{4} \)

Thus, the possible values of \( a \) such that the inequality \( 3 - x^2 > |x - a| \) has at least one negative solution are:
\[ -\frac{13}{4} \leq a \leq \frac{13}{4} \]

### Final Answer
The possible values of \( a \) are:
\[ a \in \left(-\frac{13}{4}, 3\right) \]